Find the Array Concatenation Value

🏠 ⬅️ ➡️

You are given a 0-indexed integer array nums.

The concatenation of two numbers is the number formed by concatenating their numerals.

  • For example, the concatenation of 15, 49 is 1549.

The concatenation value of nums is initially equal to 0. Perform this operation until nums becomes empty:

  • If there exists more than one number in nums, pick the first element and last element in nums respectively and add the value of their concatenation to the concatenation value of nums, then delete the first and last element from nums.
  • If one element exists, add its value to the concatenation value of nums, then delete it.

Return the concatenation value of the nums.

Example 1:

Input: nums = [7,52,2,4] Output: 596 Explanation: Before performing any operation, nums is [7,52,2,4] and concatenation value is 0. - In the first operation: We pick the first element, 7, and the last element, 4. Their concatenation is 74, and we add it to the concatenation value, so it becomes equal to 74. Then we delete them from nums, so nums becomes equal to [52,2]. - In the second operation: We pick the first element, 52, and the last element, 2. Their concatenation is 522, and we add it to the concatenation value, so it becomes equal to 596. Then we delete them from the nums, so nums becomes empty. Since the concatenation value is 596 so the answer is 596.

Example 2:

Input: nums = [5,14,13,8,12] Output: 673 Explanation: Before performing any operation, nums is [5,14,13,8,12] and concatenation value is 0. - In the first operation: We pick the first element, 5, and the last element, 12. Their concatenation is 512, and we add it to the concatenation value, so it becomes equal to 512. Then we delete them from the nums, so nums becomes equal to [14,13,8]. - In the second operation: We pick the first element, 14, and the last element, 8. Their concatenation is 148, and we add it to the concatenation value, so it becomes equal to 660. Then we delete them from the nums, so nums becomes equal to [13]. - In the third operation: nums has only one element, so we pick 13 and add it to the concatenation value, so it becomes equal to 673. Then we delete it from nums, so nums become empty. Since the concatenation value is 673 so the answer is 673.

Constraints:

  • 1 <= nums.length <= 1000
  • 1 <= nums[i] <= 104

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Note: This problem is from LeetCode.
✅ Compiled
❌ Executed
❌ Correct 
program concatenation_of_numbers
    implicit none

    integer :: i, j, n, nums(1000), concat_val

    ! read input
    read(*,*) n
    do i = 1, n
        read(*,*) nums(i)
    end do

    ! solve problem
    concat_val = 0
    do while (n > 0)
        if (n >= 2) then
            ! concatenate first and last elements
            concat_val = concat_val + concatenate(nums(1), nums(n))
            ! remove first and last elements
            nums(1:n-1) = nums(2:n)
            n = n - 1
        else
            ! concatenate single element
            concat_val = concat_val + nums(1)
            ! remove single element
            nums(1:n-1) = nums(2:n)
            n = n - 1
        end if
    end do

    ! print result
    write(*,*) concat_val

contains

function concatenate(a, b) result(concat)
    implicit none

    integer, intent(in) :: a, b
    integer :: concat

    concat = a * 10 + b
end function concatenate

end program concatenation_of_numbers
❌ Compiled
❌ Executed
❌ Correct 
PROGRAM concatenation
      IMPLICIT NONE

      INTEGER, DIMENSION(:), ALLOCATABLE :: nums
      INTEGER :: n, i, j, concatenation_value

      ! Read the input
      READ (*,*) n
      ALLOCATE (nums(n))
      DO i = 1, n
         READ (*,*) nums(i)
      END DO

      ! Initialize the concatenation value
      concatenation_value = 0

      ! Loop until the nums array is empty
      DO WHILE (SIZE(nums) > 0)
         ! Check if there are more than one number in the nums array
         IF (SIZE(nums) > 1) THEN
            ! Pick the first and last element in the nums array
            i = 1
            j = SIZE(nums)

            ! Add the concatenation of the first and last element to the concatenation value
            concatenation_value = concatenation_value + CONCATENATE(nums(i), nums(j))

            ! Delete the first and last element from the nums array
            nums(i:j-1) = nums(i+1:j)
         ELSE
            ! Add the value of the only element in the nums array to the concatenation value
            concatenation_value = concatenation_value + nums(1)

            ! Delete the only element from the nums array
            nums = nums(2:)
         END IF
      END DO

      ! Output the concatenation value
      WRITE (*,*) concatenation_value

      END PROGRAM concatenation

      FUNCTION CONCATENATE(a, b) RESULT(c)
      IMPLICIT NONE

      INTEGER, INTENT(IN) :: a, b
      INTEGER :: c

      ! Concatenate the two numbers
      c = a*10 + b

      END FUNCTION CONCATENATE
🌐 Data from online sources 
def concatenationValue(nums):
    value = 0
    while nums:
        n = len(nums)
        if n == 1:
            value += nums[0]
            nums.pop()
        else:
            value += int(str(nums[0]) + str(nums[-1]))
            nums.pop(0)
            nums.pop()
    return value

The algorithm goes as follows: 1. Initialize the concatenation value to 0. 2. Check if the nums list is empty. If it is not: a. Check if the length of the nums list is 1. If it is, add the only element in the list to the concatenation value and remove it from the list. b. If there is more than one element in the nums list, concatenate the first and last elements and add this value to the concatenation value. Then, remove both elements from the list. 3. Repeat step 2 until the nums list is empty. 4. Return the concatenation value.

🌐 Data from online sources 
int concatenationValue(vector<int>& nums) {
    int value = 0;
    while (!nums.empty()) {
        int n = nums.size();
        if (n == 1) {
            value += nums[0];
            nums.pop_back();
        } else {
            value += stoi(to_string(nums[0]) + to_string(nums[n - 1]));
            nums.erase(nums.begin());
            nums.pop_back();
        }
    }
    return value;
}

The algorithm goes as follows: 1. Initialize the concatenation value to 0. 2. Check if the nums list is empty. If it is not: a. Check if the length of the nums list is 1. If it is, add the only element in the list to the concatenation value and remove it from the list. b. If there is more than one element in the nums list, concatenate the first and last elements and add this value to the concatenation value. Then, remove both elements from the list. 3. Repeat step 2 until the nums list is empty. 4. Return the concatenation value.